- #1
cdsi385
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Does anyone know how to take the partial derivative of a convolution integral where the derivative is taken with respect to one of the functions of the convolution integral?
In the following example, the best I can come up with is:
[itex]\frac{\partial}{\partial g(t)}\int L(t-\tau)g(t)\,d\tau=\int L(t-\tau)\,d\tau[/itex]
Is this correct, or does it even make sense?
To put this in context, what I usually do (successfully) is perform the convolution integral in a simulation (without the partial differentiation) where [itex]L(t)[/itex] is the impulse response function of a system and [itex]g(t)[/itex] is the velocity of my system which is calculated on the fly during the simulation.
What I'm trying to do now is make a new simulation which relies on this partial derivative which I'm trying to express analytically before simulating it. If what I've expressed above is correct then all I need to simulate is: [itex]\int L(t-\tau)\,d\tau[/itex]
Thanks in advance...
cdsi385
In the following example, the best I can come up with is:
[itex]\frac{\partial}{\partial g(t)}\int L(t-\tau)g(t)\,d\tau=\int L(t-\tau)\,d\tau[/itex]
Is this correct, or does it even make sense?
To put this in context, what I usually do (successfully) is perform the convolution integral in a simulation (without the partial differentiation) where [itex]L(t)[/itex] is the impulse response function of a system and [itex]g(t)[/itex] is the velocity of my system which is calculated on the fly during the simulation.
What I'm trying to do now is make a new simulation which relies on this partial derivative which I'm trying to express analytically before simulating it. If what I've expressed above is correct then all I need to simulate is: [itex]\int L(t-\tau)\,d\tau[/itex]
Thanks in advance...
cdsi385